Đáp án:
$M_{MIN}=-\dfrac{1}{2}$ $\text{khi:}$
`(x; y)=(-\frac{\sqrt{3}}{2}; \frac{1}{2}); (\frac{\sqrt{3}}{2}; -\frac{1}{2})`
$M_{MAX}=\dfrac{3}{2}$ $\text{khi:}$
`(x; y)=(\frac{1}{2}; \frac{\sqrt{3}}{2}); (-\frac{1}{2}; -\frac{\sqrt{3}}{2})`
Giải thích các bước giải:
$M=\sqrt{3}xy+y^2$
$*)$ $2M+1=2\sqrt{3}xy+2y^2+x^2+y^2=x^2+2\sqrt{3}xy+3y^2=(x+\sqrt{3}y)^2$
$\text{Vì $(x+\sqrt{3}y)^2 \geq 0$ nên $2M+1 \geq 0 ⇔ M \geq -\dfrac{1}{2}$}$
$\text{Dấu "=" xảy ra khi $x=-\sqrt{3}y$}$
$⇒ (-\sqrt{3}y)^2+y^2=1$
$⇔ 4y^2=1$
$⇔ y^2=\dfrac{1}{4}$
$⇒ \left[ \begin{array}{l}y=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{array} \right.$
$⇒ \left[ \begin{array}{l}x=-\dfrac{\sqrt{3}}{2}\\x=\dfrac{\sqrt{3}}{2}\end{array} \right.$
$*)$ $2M-3=2\sqrt{3}xy+2y^2-3x^2-3y^2=-3x^2+2\sqrt{3}xy-y^2$
$=-(\sqrt{3}x-y)^2$
$\text{Vì $-(\sqrt{3}x-y)^2 \leq 0$ nên $2M-3 \leq 0 ⇔ M \leq \dfrac{3}{2}$}$
$\text{Dấu "=" xảy ra khi $\sqrt{3}x=y$}$
$⇒ x^2+(\sqrt{3}x)^2=1$
$⇔ 4x^2=1$
$⇔ x^2=\dfrac{1}{4}$
$⇒ \left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{array} \right.$
$⇒ \left[ \begin{array}{l}y=\dfrac{\sqrt{3}}{2}\\y=-\dfrac{\sqrt{3}}{2}\end{array} \right.$
$\text{Vậy $M_{MIN}=-\dfrac{1}{2}$ khi:}$
`(x; y)=(-\frac{\sqrt{3}}{2}; \frac{1}{2}); (\frac{\sqrt{3}}{2}; -\frac{1}{2})`
$M_{MAX}=\dfrac{3}{2}$ $\text{khi:}$
`(x; y)=(\frac{1}{2}; \frac{\sqrt{3}}{2}); (-\frac{1}{2}; -\frac{\sqrt{3}}{2})`
